Very hard Physics question

  • Thread starter mack2629
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  • #1
mack2629
Very hard Physics question!!

This question is very hard, a football is kicked at an 50 degree angle to the horizontal , travels a horizontal distance of 20m before hitting the ground. what is the initial speed?
please tell me how you would do this
 

Answers and Replies

  • #2
Tom Mattson
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Hi Mack, welcome to PF.

Originally posted by mack2629
This question is very hard,
No, it is not.

please tell me how you would do this
Sorry, but I have a policy here: You show us how you start and where you get stuck, and we help you through the rough spots. I have it stuck to the top of the forum: "Read This Before Posting". We don't do yer homework for you here.

This is a basic projectile motion problem. Start by writing down your knowns and unknowns, and--most importantly--read up on projectile motion.
 
  • #3
333
1
If you have read the rules of this forum, we are not able to answer your question completely.
But i will give you some clues.
From these three equations (you may not need them all), you can derive an equation that makes a relation between the horizontal distance travelled, the angle of the kick, and the initial speed (if you already know this equation just use it).
After you get the equation, plug in what you know (the distance and angle), and solve for the unknown (the intial speed).

(you will have to apply the equations on the horizontal movement alone, and the vertical movement alone)
S=vi*t+0.5*a*t2
vf2=vi2+2*a*S
vf=a*t+vi

If you are still stuck, tell us and we will help you a little more .
 
  • #4
mack2629
i know, V1x=Vcos50
Ax=0m/s*
dx=20m
t=?

V1y=Vsin50
Ay=-9.8m/s*
dy=?
t=?

t=20/Vcos50
V1x=20/t
and im stuckafter this because when i plug my first equation in to the second you cant solve it.

by the way im in grade 11.
 
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  • #5
petitbear
Hint: You might want to use Vsin50 and Vcos50 instead of Vy and Vx. (1)g=Vsin50*(t/2) (2)s=Vcos50*t

Do you know, how (1) is formed?
 
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  • #6
333
1
Okz, here are some little tricks.
Since you are omiting the friction with air, then the energy will be reserved.
Since the energy is reserved, the speed of the object at launch will be exactly equal to the speed of the object when it hist back the ground (but opposite in direction).
So ..
vf=a*t+vi
But vf=-vi
-vi=a*t+vi
-2vi=a*t
t=-2vi/a

Now you can use that equation on the vertical movement to get the time of the whole trip.

S=vi*t+0.5*a*t2
Apply this on the horizontal movement, note that that the only unknowns are vi and t.
Subsitute the value of t from the first equation.
Now you have a single equation (quadratice) with a single variable, solve it.

Still stuck ?
 
  • #7
emu
The only formula you can use is the one for range. I'm not sure if you've even seen it yet.
At any rate,
Range = V^2 * sin(2*Theta) / g
 
  • #9
508
0
Hi mack2629, I believe that it always helps to first figure out what is happening.
1) No friction -> path of flight is parabola. Launch speed is equal to impact speed (!)
2) Zeroes of parabola are 0 and 20, so y = ax(x-20).
3) 50 degree angle of launch, so dy/dx = tan 50 when x = 0.
4) Calculus: dy/dx = 2ax-20a, so -20a = tan 50.
Now figure out how high is the highest point of that parabola, and what time it takes to fall from that height, and what speed is reached when hitting the ground after having fallen from that height, and use a little common sense and some Pythagoras, and you 've got it. :wink:
 
  • #10
1
0


range=(sin(2a)u^2)/g
 
  • #11
409
0


range=(sin(2a)u^2)/g
Well done you just necroposted in a thread that is 8 years, 9 months, 2 weeks, 1 day old.
 

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